Hello!

Please, what should I do with the sqrt in the denominator?

Please, help someone!!!!

Many thanks!

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- Jan 27th 2013, 03:33 AMBoointegral with sqrt in denominator
Hello!

Please, what should I do with the sqrt in the denominator?

Please, help someone!!!!

Many thanks! - Jan 27th 2013, 04:27 AMibduttRe: integral with sqrt in denominator
Workout the denominator in the form ∫▒〖1/√(a^2-x^2 ) OR 〗 ∫▒〖1/√(a^2+x^2 ) 〗and then use the formula

∫▒〖1/√(a^2-x^2 ) dx = sin^(-1)〖x/a+C OR 〗 ∫▒〖1/√(a^2+x^2 ) 〗 dx= log〖|x+ 〗 √(a^2+x^2 )|+C〗

2 + 3x – 2x2 = 2 – 2[ x2 - 3/2 x] = 2 – 2[ x2 - 3/2 x + 〖(3/4)〗^2- 〖(3/4)〗^2]

= 2 – 2[ 〖( x- 3/4)〗^2- 〖(3/4)〗^2]

= 2[ 1 – 〖( x- 3/4)〗^2+ 9/16]

= 2[25/16 – 〖( x- 3/4)〗^2]

Etc… - Jan 27th 2013, 04:38 AMILikeSerenaRe: integral with sqrt in denominator
Well, let's see if we can find a known anti-derivative for an expression with a square root in the denominator.

Hmm, we know that the anti-derivative of is .

This looks a bit like it.

Can you perhaps write the argument of the square root in the form ?

After that you can make the substitution u=bx-c. - Jan 27th 2013, 05:11 AMBooRe: integral with sqrt in denominator
Thank u, both!!!

Well, 2+3x-2x^2=1-(bx-c)^2

1+3x-2x^2=-(bx-c)^2= - (b^2x^2-2bc+c^2)

=-b^2+2bc-c^2

dont see the soution: -c^2should be 1 then... - Jan 27th 2013, 05:18 AMProve ItRe: integral with sqrt in denominator
- Jan 27th 2013, 05:21 AMBooRe: integral with sqrt in denominator
I shoudl solve it like this:

2+3x-2x^2= (25/8) (1-(x-3/4)^2/(25/16)

and put in:

t= 4(x-3/4)/5

huh...any help??? - Jan 27th 2013, 05:32 AMILikeSerenaRe: integral with sqrt in denominator
You have left out the factor a...

But let's see:

So... substitute:

- Jan 28th 2013, 07:25 AMBooRe: integral with sqrt in denominator
Please how did u get:

- Jan 28th 2013, 11:25 AMILikeSerenaRe: integral with sqrt in denominator
- Jan 28th 2013, 03:37 PMProve ItRe: integral with sqrt in denominator
Experience. Basically when you see something of the form in the denominator and you can't do a u-substitution, then a substitution of is appropriate, as it will simplify using the Pythagorean Identity and then cancel with its own derivative of , making the integration unbelievably easy.

- Jan 28th 2013, 03:49 PMILikeSerenaRe: integral with sqrt in denominator
- Jan 28th 2013, 03:52 PMProve ItRe: integral with sqrt in denominator
- Jan 29th 2013, 04:13 AMBooRe: integral with sqrt in denominator
PLEASE; HOW DID U GET .

?

Many Thanks-stuck at basic things!!! - Jan 29th 2013, 06:11 AMProve ItRe: integral with sqrt in denominator
Like I told you, it comes with experience. Let's just take a look at the arbitrary integral .

Notice that it looks extremely complicated. One might think that since there is a sine function in the integrand and the cosine function (sine's derivative) is a multiple, it might be appropriate to do a substitution. So if we let , then we would have

But this does NOT help us (in fact, it gets us to almost the exact same integral that we were trying to find in your original post!) so a different tactic is required.

You should be well aware of the Pythagorean Identity in trigonometry: . We can use this fact to help us simplify the integrand. Notice that this means . Can you see that we have almost this exact quantity in the denominator of the starting integral in this post?

which is an EXTREMELY easy quantity to integrate.

So that means that when you have something of the form or similar, it is appropriate to use the substitution due to the fact that the denominator simplifies to a cosine function by Pythagoras, and you also end up with a cosine function in the derivative which it cancels with, giving you an extremely easy quantity (such as a constant) to integrate.